Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 28}{x + 6} = \dfrac{x + 118}{x + 6}$
Answer: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 28}{x + 6} (x + 6) = \dfrac{x + 118}{x + 6} (x + 6)$ $ x^2 + 28 = x + 118$ Subtract $x + 118$ from both sides: $ x^2 + 28 - (x + 118) = x + 118 - (x + 118)$ $ x^2 + 28 - x - 118 = 0$ $ x^2 - 90 - x = 0$ Factor the expression: $ (x - 10)(x + 9) = 0$ Therefore $x = 10$ or $x = -9$ The original expression is defined at $x = 10$ and $x = -9$, so there are no extraneous solutions.